Question

a ball is dropped from a height h on a horizontal floor if it loses 60% of its energy on heating the floor then the height up to which it will rise after first rebounds is​

Answer 1

Answer:

May be 2h/5 is the correct answer

Explanation:

When body falls from height h its

potential energy = mgh

initially K.E = P.E

mgh = $\frac{mv^{2} }{2}$

$v = \sqrt{2gh}$

after hitting the ground it losses 60% but still it has 40% of its initial energy

40% of  $\frac{mv^{2} }{2}$ = $\frac{mv'^{2} }{2}$

$v'^{2}  = \frac{2v^{2} }{5}$

so, $\frac{mv'^{2} }{2} = mgh'$

put the value of v'^2    and we get

h' = 2h/5